Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.1 N when they are separated by 60.9 cm. What is the magnitude of the charges in microCoulombs?

Answer 1

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

`\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}*10^(-2)\text{ m} \end{gathered}`

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Step-by-step explanation:

The magnitude of charge can be calculated by the formula

`\begin{gathered} F=(k(2q))/(r^2) \\ q=(Fr^2)/(2k) \end{gathered}`

Here, k is the Coulomb's constant whose value is

`k\text{ = 9}*10^9\text{ N m}^2\text{ /C}^2`

On substituting the values, the magnitude of charge will be

`\begin{gathered} q=(48.1*(60.9^*10^(-2))^2)/(2*9*10^(^9)) \\ =9.91*10^(-10)\text{ C} \\ =9.91*10^(-4)\text{ }\mu C \end{gathered}`

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.